You have found the following ages (in years) of all 6 turtles at your local zoo: $ 55,\enspace 95,\enspace 2,\enspace 36,\enspace 13,\enspace 93$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 6 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{55 + 95 + 2 + 36 + 13 + 93}{{6}} = {49\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $55$ years $6$ years $36$ years $^2$ $95$ years $46$ years $2116$ years $^2$ $2$ years $-47$ years $2209$ years $^2$ $36$ years $-13$ years $169$ years $^2$ $13$ years $-36$ years $1296$ years $^2$ $93$ years $44$ years $1936$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{36} + {2116} + {2209} + {169} + {1296} + {1936}} {{6}} $ $ {\sigma^2} = \dfrac{{7762}}{{6}} = {1293.67\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{1293.67\text{ years}^2}} = {36\text{ years}} $ The average turtle at the zoo is 49 years old. There is a standard deviation of 36 years.